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2 years ago

I need help plzzzzzz

Chemistry

2 answers:

Vladimir [108]2 years ago

4 0

I think 25%?
I hope this helps

tigry1 [53]2 years ago

4 0

Answer:

it is 25%

Explanation:

Answering so you can give the other person brainliest :D have a good day

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The term for free electrons in metals transferring energy by colliding with other atoms or electrons is _____.

shepuryov [24]

Conduction

In metals, the electrons of each atom are delocalized, which means they are free to move about in the structure of the metal. When heat is applied at one end of the metal, the electrons there are excited and collide with surrounding particles, transferring their energy to them. This is what makes metals very good conductors of both heat and electricity.

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3 years ago

Read 2 more answers

Distinguish between a carbohydrate and a protein

Verdich [7]

A carbohydrate is a trans fat while protien is nor

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3 years ago

A solution contains 3.1 mM Zn(NO3)2 and 4.2 mM Ca(NO3)2. The p-function for Zn2+ is _____, and the p-function for NO3- is _____.

seropon [69]

Answer: The p-function of $Zn^{2+}$ and $NO_3^{-}$ ions are 2.51 and 2.14 respectively.

Explanation:

p-function is defined as the negative logarithm of any concentration.

We are given:

Millimolar concentration of zinc nitrate = 3.1 mM

Millimolar concentration of calcium nitrate = 4.2 mM

Converting this into molar concentration, we use the conversion factor:

1 M = 1000 mM

Concentration of zinc nitrate = 0.0031 M = 0.0031 mol/L

1 mole of zinc nitrate produces 1 mole of zinc ions and 2 moles of nitrate ions

Concentration of zinc ions = 0.0031 M

Concentration of nitrate ions in zinc nitrate, $M_1=(2\times 0.0031)=0.0062M$

Concentration of calcium nitrate = 0.0042 M = 0.0042 mol/L

1 mole of calcium nitrate produces 1 mole of calcium ions and 2 moles of nitrate ions

Concentration of calcium ions = 0.0042 M

Concentration of nitrate ions in calcium nitrate, $M_2=(2\times 0.0042)=0.0084M$

To calculate the concentration of nitrate ions in the solution, we use the equation:

$M=\frac{M_1V_1+M_2V_2}{V_1+V_2}$

Putting values in above equation, we get:

$M=\frac{(0.0062\times 1)+(0.0084\times 1)}{1+1}\\\\M=0.0073M$

Calculating the p-function of zinc ions and nitrate ions in the solution:

For zinc ions:

$\text{p-function of }Zn^{2+}\text{ ions}=-\log[Zn^{2+}]$

$\text{p-function of }Zn^{2+}\text{ ions}=-\log(0.0031)\\\\\text{p-function of }Zn^{2+}\text{ ions}=2.51$

For nitrate ions:

$\text{p-function of }NO_3^{-}\text{ ions}=-\log[NO_3^{-}]$

$\text{p-function of }NO_3^{-}\text{ ions}=-\log(0.0073)\\\\\text{p-function of }NO_3^{-}\text{ ions}=2.14$

Hence, the p-function of $Zn^{2+}$ and $NO_3^{-}$ ions are 2.51 and 2.14 respectively.

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3 years ago

What is a zero order reaction? Give two examples.

dezoksy [38]

Answer:

I guess 3 to be honest and that's what I think

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3 years ago

Predict whether ΔS° is greater than, less than, or approximately zero for each of the following reactions, and explain your choi

inna [77]

Answer:

Explanation:

Entropy -

In a system, the randomness is measured by the term entropy .

Randomness basically refers as a form of energy that can not be used for any work.

The change in entropy is given by amount heat per change in temperature.

When solid is converted to gas entropy increases,

As the molecules in solid state are tightly packed and has more force of attraction between the molecules, but as it is converted to gas, the force of attraction between the molecule decreases and hence entropy increases.

So,

The particles of the substance , if are tightly held by strong force of attraction will decrease the entropy ,

And

If the particles are loosely held , the entropy will increase .

If in a reaction , more number of gaseous atoms are present in the product side , entropy will increase , i.e. Δ°S > 0

When liquid is converted to solid entropy decreases,

As the molecules in liquid state are loosely packed and has less force of attraction between the molecules, but as it is converted to solid, the force of attraction between the molecule increases and hence entropy decreases.

If in a reaction , less number of gaseous atoms are present in the product side , entropy will decrease , i.e. Δ°S < 0

From the question ,

( a ) NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)

Gaseous atoms -

Reactant - 1 + 5 = 6

Product - 4 + 6 = 10 ,

Hence ,

More number of gaseous atoms are present in the product side , So ,

entropy will increase , i.e. Δ°S > 0

( b ) CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g)

Gaseous atoms -

Reactant - 1 + 2 = 3

Product - 1 + 2 = 3 ,

Since ,

Both the side the value of gaseous atoms are , hence , Δ°S = 0 .

( c ) CaCO₃(s) → CaO(s) + CO₂(g)

Gaseous atoms -

Reactant = 0

Product - 0 + 1 = 1 ,

Since ,

Hence ,

More number of gaseous atoms are present in the product side , So ,

entropy will increase , i.e. Δ°S > 0

8 0

3 years ago

I need help plzzzzzz​

I need help plzzzzzz